# HOW TO EVALUATE INTEGRALS USING SUBSTITUTION

Key Idea :

Let us consider the integration of a function with respect to x. If the derivative of a part of the function is at somewhere of the function and you are able to multiply the derivative along with dx using multiplication, then you can integration the function using substitution.

Consider the following integration.

In the function the power of 'e' is x3. The derivative is x3 is 3x2 and it is being as a part of the function and we will be able to write the 3x2 along with dx using multiplication.

Now substitute a new variable for x3 for which we have the derivative 3x2 being a part of the given function.

Let u = x3.

u = x3

Differentiate with respect to x on both sides.

du/dx = 3x2

Multiply both sides by dx.

du = 3x2dx

In the given integration, substitute x3 = u and 3x2dx = du.

= ∫eudu

= eu + C

Substitute u = x3.

Question 1 :

Integrate the following with respect to x

cot x/log (sin x)

Solution :

=  ∫cot x/log (sin x) dx

Let u = log (sin x)

Differentiating with respect to "x"

du/dx  =  1/sin x ( cos x)

du  =  cos x / sin x

du  =  cot x dx

∫cot x/log (sin x) dx  =  ∫du/u

=  log u + c

=  log (log (sin x)) + c

Question 2 :

Integrate the following with respect to x

cosec x/log (tan (x/2))

Solution :

=  ∫ [cosec x/log (tan (x/2))] dx

Let u  =  log (tan x/2)

du  =  (1/tan (x/2)) ⋅ sec2 (x/2) (1/2) dx

=  (cos x/2 / sin x/2) ⋅ (1/cos2 (x/2)) (1/2) dx

=  (1 / 2 sin x/2 cos (x/2))dx

=  1/ sin x dx

du  =   cosec x dx

∫ [cosec x/log (tan (x/2))] dx   =  ∫ du/u

=  log u + c

=  log (log tan x/2) + c

Question 3 :

Integrate the following with respect to x

sin 2x / (a2 + b2 sin2x)

Solution :

=  ∫ [sin 2x / (a2 + b2 sin2x)] dx

Let u  =  a2 + b2 sin2x

du  =  0 + b2 (2 sin x cos x) dx

du  =   b2 sin 2x dx

sin 2x dx  =  du/b2

∫ [sin 2x / (a2 + b2 sin2x)] dx  =  ∫ (du/b2)/u

=  (1/b2) ∫ (1/u) du

=  (1/b2) log u + c

=  (1/b2) log (a2 + b2 sin2x) + c

Question 4 :

Integrate the following with respect to x

sin-1x/√(1 - x2)

Solution :

=  ∫ [sin-1x/√(1 - x2)] dx

Let u = sin-1x

du  =  (1/√(1 - x2)) dx

∫ [sin-1x/√(1 - x2)] dx  =  ∫ u du

=  u2/2 + c

=  (sin-1x)2/2 + c

Question 5 :

Integrate the following with respect to x

√x/(1 + √x)

Solution :

=  ∫ [√x/(1 + √x)] dx

Let u = 1 + √x

du  =  (0 + 1/2√x) dx

2√x du  =  dx

∫ [√x/(1 + √x)] dx  =  ∫ [ (u - 1)/u] 2 √x du

=  ∫ [ (u - 1)/u] 2 (u - 1) du

=  2 ∫ [ (u - 1)2/u] du

=  2 ∫ [u2 - 2u + 1)/u] du

=  2 [(∫u du) - 2 ∫ du + ∫(1/u) du]

=  2 [u2/2 - 2u + log u] + c

=  u- 4u + 2log u + c

(1 + √x)- 4(1 + √x) + 2log (1 + √x) + c

Question 6 :

Integrate the following with respect to x

1/(x log x log (log x))

Solution :

=  ∫ [1/(x log x log (log x))] dx

Let u  =  log (log x)

du  =  (1/log x) (1/x)

du  =  (1/x log x) dx

∫ [1/(x log x log (log x))] dx = 1/(x logx) (log (logx))] dx

=  ∫ (du/u)

=  log u

=  log (log (log x)) + c

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